Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int MAXN = 1e5 + 10;
void add(long long &x, long long y) {
  x += y;
  x %= MOD;
}
struct Tree {
  int l, r;
  int Q[10], cnt;
};
Tree tree[MAXN << 2];
void PushUp(int o) {
  int os = 0, ls = tree[o << 1].cnt, rs = tree[o << 1 | 1].cnt;
  int lc = 0, rc = 0;
  while (os < 10) {
    if (lc < ls && rc < rs) {
      if (tree[o << 1].Q[lc] > tree[o << 1 | 1].Q[rc]) {
        tree[o].Q[os++] = tree[o << 1 | 1].Q[rc++];
      } else {
        tree[o].Q[os++] = tree[o << 1].Q[lc++];
      }
    } else if (lc < ls) {
      tree[o].Q[os++] = tree[o << 1].Q[lc++];
    } else if (rc < rs) {
      tree[o].Q[os++] = tree[o << 1 | 1].Q[rc++];
    } else
      break;
  }
  tree[o].cnt = os;
}
priority_queue<int, vector<int>, greater<int> > QQ[MAXN];
void Build(int o, int l, int r) {
  tree[o].l = l;
  tree[o].r = r;
  tree[o].cnt = 0;
  if (l == r) {
    return;
  }
  int mid = (l + r) >> 1;
  Build(o << 1, l, mid);
  Build(o << 1 | 1, mid + 1, r);
}
void Update(int o, int index, int pos) {
  if (tree[o].l == tree[o].r) {
    int os = 0;
    while (!QQ[index].empty() && os < 10) {
      tree[o].Q[os++] = QQ[index].top();
      QQ[index].pop();
    }
    tree[o].cnt = os;
    return;
  }
  int mid = (tree[o].l + tree[o].r) >> 1;
  if (pos <= mid)
    Update(o << 1, index, pos);
  else
    Update(o << 1 | 1, index, pos);
  PushUp(o);
}
int ans[10], ans1[10], res;
void Query(int o, int L, int R) {
  if (tree[o].l == L && tree[o].r == R) {
    memcpy(ans1, ans, sizeof(ans));
    int os = 0, ls = res, rs = tree[o].cnt;
    int lc = 0, rc = 0;
    while (os < 10) {
      if (lc < ls && rc < rs) {
        if (tree[o].Q[rc] > ans1[lc]) {
          ans[os++] = ans1[lc++];
        } else {
          ans[os++] = tree[o].Q[rc++];
        }
      } else if (lc < ls) {
        ans[os++] = ans1[lc++];
      } else if (rc < rs) {
        ans[os++] = tree[o].Q[rc++];
      } else
        break;
    }
    res = os;
    return;
  }
  int mid = (tree[o].l + tree[o].r) >> 1;
  if (R <= mid)
    Query(o << 1, L, R);
  else if (L > mid)
    Query(o << 1 | 1, L, R);
  else {
    Query(o << 1, L, mid), Query(o << 1 | 1, mid + 1, R);
  }
}
struct Edge {
  int from, to, next;
};
Edge edge[MAXN << 1];
int head[MAXN], edgenum;
void init() {
  edgenum = 0;
  memset(head, (-1), sizeof(head));
}
void addEdge(int u, int v) {
  Edge E = {u, v, head[u]};
  edge[edgenum] = E;
  head[u] = edgenum++;
}
int son[MAXN], num[MAXN];
int top[MAXN], pos[MAXN], id;
int dep[MAXN], pre[MAXN];
void DFS1(int u, int fa, int d) {
  dep[u] = d;
  pre[u] = fa;
  num[u] = 1;
  son[u] = -1;
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].to;
    if (v == fa) continue;
    DFS1(v, u, d + 1);
    num[u] += num[v];
    if (son[u] == -1 || num[son[u]] < num[v]) son[u] = v;
  }
}
void DFS2(int u, int T) {
  top[u] = T;
  pos[u] = ++id;
  if (son[u] == -1) return;
  DFS2(son[u], T);
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].to;
    if (v == pre[u] || v == son[u]) continue;
    DFS2(v, v);
  }
}
void Solve(int u, int v) {
  int f1 = top[u], f2 = top[v];
  res = 0;
  while (f1 != f2) {
    if (dep[f1] < dep[f2]) {
      swap(u, v);
      swap(f1, f2);
    }
    Query(1, pos[f1], pos[u]);
    u = pre[f1], f1 = top[u];
  }
  if (dep[u] > dep[v]) swap(u, v);
  Query(1, pos[u], pos[v]);
}
int main() {
  int n, m, q;
  while (scanf("%d%d%d", &n, &m, &q) != EOF) {
    init();
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      addEdge(u, v);
      addEdge(v, u);
    }
    DFS1(1, -1, 1);
    id = 0;
    DFS2(1, 1);
    Build(1, 1, id);
    for (int i = 1; i <= m; i++) {
      int v;
      scanf("%d", &v);
      if (QQ[v].size() < 10) {
        QQ[v].push(i);
      }
    }
    for (int i = 1; i <= n; i++) {
      Update(1, i, pos[i]);
    }
    while (q--) {
      int u, v, k;
      scanf("%d%d%d", &u, &v, &k);
      Solve(u, v);
      int temp = min(k, res);
      printf("%d", temp);
      for (int i = 0; i < temp; i++) {
        printf(" %d", ans[i]);
      }
      printf("\n");
    }
  }
  return 0;
}